Question: Suppose we have a vector-valued function $g(t)$ and a scalar function $f(x, y)$. Let $h(t) = f(g(t))$. We know: $\begin{aligned} &g(-2) = (0, 0) \\ \\ &g'(-2) = (-2, 9) \\ \\ &\nabla f(0, 0) = (4, 1) \end{aligned}$ Evaluate $\dfrac{d h}{d t}$ at $t = -2$. $h'(-2)=$
Answer: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(-2) = \nabla f(g(-2)) \cdot g'(-2)$. We know the following. $\begin{aligned} &g(-2) = (0, 0) \\ \\ &g'(-2) = (-2, 9) \\ \\ &\nabla f(0, 0) = (4, 1) \end{aligned}$ Substituting: $h'(-2) = (4, 1) \cdot (-2, 9) = 1$ Answer Therefore, $h'(-2) = 1$.